Past exam of the mathematics course of the University of Cambridge 2017 ii Paper 2 16I c Solution Created 2026-09-24 Updated 2026-10-05
In characteristic , let and , with having algebraic independence. The monomials , , form a -basis, so . Their independence follows by grouping polynomial monomials according to their exponents modulo , after clearing rational-function denominators.
Every has , so its minimal polynomial divides and . Thus this finite purely inseparable extension has no primitive element. It demonstrates why the separable field extension hypothesis cannot simply be removed.
Past exam of the mathematics course of the University of Cambridge 2018 iii Paper 139 3 iv Solution Created 2026-10-03 Updated 2026-10-05
Choose a very ample divisor . By the stronger form of Kodaira's lemma, for some effective divisor . The subsystem defines the embedding given by on . Ratios of its sections generate the function field . The map from the complete complete linear system of a divisor contains these ratios, so it induces the same function field and is birational onto its image.
The converse is true; smoothness is unnecessary. If gives a birational map, choose algebraically independent ratios among a generating set of its section ratios. For every , the sectionsare linearly independent by algebraic independence. Therefore , giving condition (1) of part (iii). This is the birational linear system criterion for bigness.